A Numerical Method to Find the Probability of Ultimate Ruin

Insurance: Mathematics and Economics 36 (2005) 399–420

A numerical method to find the probability of ultimate ruin in theclassical risk model with stochastic return on investments

Jostein Paulsena,∗, Juma Kasozib, Andreas Steigenc

a Department of Mathematics, University of Bergen, Johs. Brunsgt. 12, 5008 Bergen, Norwayb Department of Mathematics, Makerere University, Kampala, Uganda

c Center of Environmental and Resource Studies, University of Bergen, Nygaardsgt. 5, 5020 Bergen, Norway

Received July 2004 ; received in revised form February 2005; accepted 11 March 2005

Abstract

Letψ(y) be the probability of ultimate ruin in the classical risk process compounded by a linear Brownian motion. Herey isthe initial capital. We give sufficient conditions for the survival probability functionφ = 1 − ψ to be four times continuouslydifferentiable, which in particular implies thatφ is the solution of a second order integro-differential equation. Transforming thisequation into an ordinary Volterra integral equation of the second kind, we analyze properties of its numerical solution whenbasically the block-by-block method in conjunction with Simpsons rule is used. Finally, several numerical examples show thatthe method works very well.© 2005 Elsevier B.V. All rights reserved.

Keywords:Ruin probability; Volterra equation; Block-by-block method

1. Introduction

The problem of finding the probability of ultimate ruin for a risk process has been the subject of a vast number ofpapers in the insurance literature. By far the majority of these papers are concentrated on the analytical aspects ofthe problem, but there is also a quite considerable number that deal with numerical methods to actually calculate thisevasive probability. These papers are roughly divided into two groups, the largest attempting to solve the relevantintegral- or integro-differential equation numerically, and the other using Monte Carlo techniques. Examples ofpapers in this latter group areLehtonen and Nyrhinen (1992)andAsmussen and Binswanger (1997)for the classicalrisk process whileAsmussen and Nielsen (1995)allows for a deterministic return on investments. However, it is

∗ Corresponding author.E-mail addresses:[email protected] (J. Paulsen), [email protected] (J. Kasozi), [email protected] (A. Steigen).

0167-6687/$ – see front matter © 2005 Elsevier B.V. All rights reserved.doi:10.1016/j.insmatheco.2005.02.008

400 J. Paulsen et al. / Insurance: Mathematics and Economics 36 (2005) 399–420

clear from the paper byPaulsen and Rasmussen (2003)that successful use of Monte Carlo techniques in the presenceof a stochastic return on investments is by no means a routine task.

When it comes to numerical methods for the classical risk process, the literature abounds with suggestions, seee.g.De Vylder (1996)for an overview of some of these methods. Surprisingly few of the suggested methods takeadvantage of the vast knowledge about integral equations in the numerical literature, and as a consequence manydo not pay much attention to the error rate inherent in any numerical method. An exception is the paper byRamsayand Usabel (1997)where the method of product integration is used to solve the Volterra integral equation for theruin probability.

It should also be mentioned that there is a considerable literature dealing with numerical methods for calculatingthe ruin probability in finite time, seeDe Vylder (1996)for several of these. More recent papers allowing fora constant return on investments areDickson and Waters (1999), Brekelmans and De Waegenaere (2001), andCardoso and Waters (2003). See also the survey paper byPaulsen (1998a). Since very little is known analyticallyhere, numerical methods are even more important than in the infinite time case.

In this paper we will be concerned with infinite time ruin probabilities for the diffusion perturbated classicalrisk process compounded by a linear Brownian motion, see Section2 for the details. When there is no Brownianmotion, the abovementioned paper byDickson and Waters (1999)presents several numerical methods from theliterature and show by numerical examples that some of them work well. However, none of these methods belongto the standard numerical methods developed by numerical analysts to solve the relevant equations, and conse-quently Dickson and Waters do not discuss the error rate of the various methods. In Section2, we use integrationby parts to represent the survival probability as a linear Volterra integral equation of the second kind, and thenuse (in one case a modification of) the standard block-by-block method in conjunction with the Simpson rule tosolve this equation. Partly relying on known results in the numerical literature, we prove in Section3 that thenumerical solution has an error of order 4. For this to hold it is necessary that the survival function is four timescontinuously differentiable, and some effort is spent in Section2 to prove this fact. We also prove in Section3 thatthe error rate of order 4 still holds when functions appearing in the kernel of the integral operator are calculatednumerically using the Simpson rule. Finally, in Section4, several numerical examples are given showing that themethod really works. Here it is also shown how known asymptotic results can be used to approximate very smallruin probabilities and also how they can be used to facilitate calculations when ruin probabilities are very heavytailed.

In this paper we concentrate on the case when assets earn return on investments. Needless to say, our methodswould of course work equally well for the simpler case with no such return.

2. The model and some theoretical results

We will assume that all processes and random variables are defined on a filtered probability space (Ω,F,F, P)satisfying the usual conditions, i.e.Ft is right continuous andP-complete.

The basic process is the surplus generating process

Pt = pt + σPWP,t −Nt∑i=1

Si, t ≥ 0. (2.1)

HereWP is a standard Brownian motion independent of the compound Poisson process∑N·

i=1 Si. We denote theintensity ofNbyλ, and letF be the distribution function of theSi. It is assumed thatF is continuous and concentratedon (0,∞). In the literatureP is called a classical risk process perturbated by a diffusion.

The interpretation of(2.1) is thatp is the premium intensity,NP is the claim number process and theSP,i arethe claim sizes. The assumptionFP (0) = 0 assures that they are positive. The Brownian termσPWP is meant totake care of small perturbations in the other two terms.

J. Paulsen et al. / Insurance: Mathematics and Economics 36 (2005) 399–420 401

The risk processY is the solution of the linear stochastic differential equation

Yt = y + Pt +∫ t

0Ys− dRs, (2.2)

whereR is the return on investment generating process

Rt = rt + σRWR,t, t ≥ 0, (2.3)

andWR is another Brownian motion, independent of the processP. The interpretation of(2.3)is thatr is the nonriskypart of the investments so thatRt = rt implies that one unit invested at time zero will be worth ert at timet. ThetermσRWR then takes care of the random fluctuations in the investment return.

By Paulsen (1993)the solution of(2.2) is

Yt = Rt

(y +

∫ t

0R−1s dPs

), (2.4)

where

Rt = exp

(r − 1

2σ2R

)t + σR WR,t

, t ≥ 0,

is just the standard geometric Brownian motion so extensively used in mathematical finance. It is the solution ofdRt = rRt dt + σRRt dWR,t with R0 = 1.

Since bothP andRhave stationary independent increments,Y is a homogeneous strong Markov process. UsingIto’s formula we find that the infinitesimal generator forY is given by

Ag(y) = 1

2(σ2Ry

2 + σ2P )g′′(y) + (ry + p)g′(y) + λ

∫ ∞

0(g(y − x) − g(y)) dF (x).

LetTy = inf t : Yt < 0 be the time of ruin, withTy = ∞ if ruin never occurs and letψ(y) = P(Ty < ∞) be theprobability of eventual ruin. The following simple, but very useful result is proved inPaulsen and Gjessing (1997)Theorem 2.1.

Theorem 2.1. Assume that the equationAφ = 0 has a bounded, twice continuously differentiable solution(onescontinuously differentiable ifσR = σP = 0) that satisfies the boundary conditions

φ(y) = 0 on y < 0,

φ(0) = 0 if σ2P > 0,

limy→∞φ(y) = 1.

(2.5)

Thenφ(y) = 1 − ψ(y) is the probability of survival.

Using thatφ(y) = 0 for y < 0, Aφ = 0 takes the form

1

2(σ2Ry

2 + σ2P )φ′′(y) + (ry + p)φ′(y) + λ

∫ y

0φ(y − x) dF (x) − λφ(y) = 0. (2.6)

Eq. (2.6) is a second order integro-differential equation of the Volterra type. However, repeated integration byparts brings it into an ordinary Volterra integral equation, which will be the object of our numerical study.


Theorem 2.2. The integro-differential equation(2.6)can be represented as the Volterra integral equation of thesecond kind

φ(y) +∫ y

0K(y, x)φ(x) dx = α(y), (2.7)

where for the case with no diffusion, i.e.σ2P = σ2

R = 0,

K(y, x) = − r + λF (y − x)

ry + p, α(y) = p

ry + pφ(0), (2.8)

with F (x) = 1 − F (x). When there is diffusion, i.e.σ2P + σ2

R > 0,

K(y, x) = 2(2r − 3σ2

R + λ)x+ p+ λF2(y − x) − (r − σ2R + λ)y

σ2Ry

2 + σ2P

,

α(y) =

2p

σ2Ryφ(0) if σ2

P = 0,

σ2Py

σ2Ry

2 + σ2P

φ′(0) if σ2P > 0,

(2.9)

with F2(x) = ∫ x0 F (v) dv.

Remark. There is an apparent singularity at 0 for the caseσ2P = 0 andσ2

R > 0, but a simple Taylor expansion ofφ around zero shows that it is a removable singularity. More details here are given in the proof ofTheorem 3.1.

Proof. Replacingy in (2.6)with zand integrating by parts w.r.t.zon [0, y] gives in combination with(2.5)

1

2(σ2Ry

2 + σ2P )φ′(y) − 1

2σ2Pφ

′(0) + ((r − σ2R)y + p)φ(y) − pφ(0) − (r − σ2

R + λ)∫ y

0φ(x) dx

+ λ

∫ y

0F (y − x)φ(x) dx = 0. (2.10)

In particular if there is no diffusion, i.e. ifσP = σR = 0, this is just(2.7) and (2.8).If either or both ofσP andσR are nonzero, then replacingy in (2.10)with zand again integrating by parts w.r.t.

zon [0, y] using(2.5)shows thatφ still satisfies the equation(2.7)with K(y, x) andα(y) given by(2.9).

Before a deeper analysis is started, it is useful to be aware of the trivial cases. A proof of the following result canbe found inPaulsen (1998b).

Theorem 2.3. Letφ(y) be the survival probability and assume thatp > 0, λ > 0 andr > 0.Thenφ(y) > 0 for ally > 0 if and only ifr > 1

2σ2R, and in this caseφ(∞) = 1.Whenr ≤ 1

2σ2R, φ(y) = 0 for all y.

Remark. The fact thatφ(y) ≡ 0 wheneverr ≤ 12σ

2R may seem a bit surprising at first sight, but it is easily explained.

Assume first thatr < 12σ

2R. ThenR−1

t → ∞ ast → ∞ at a geometric rate, so thatVt = ∫ t0 R

−1s dPs will fluctuate

more and more. Now by(2.4), Ty = inf t : Vt < −y, and because of the large fluctuations,Ty will with probabilityone be finite. The case withr = 1

2σ2R is the limiting case, givingR−1

t = e−σRWR,t . Therefore, the oscillatory nature

of WR again assures that ruin will happen. Whenr > 12σ

2R, thenR−1

t → 0 ast → 0 at a geometric rate, so thatVt → V∞, whereV∞ is a finite random variable. So in this caseTy may or may not be a finite.


From now on it is assumed thatr > 12σ

2R. To return to our Eq.(2.8), fix g(0) (org′(0) if σ2

P > 0) and consider

g(y) +∫ y

0K(y, x)g(x) dx = β(y), (2.11)

whereβ(y) is asα(y) in (2.9) or (2.10), but withφ(0)(φ′(0)) replaced byg(0)(g′(0)). If g is a bounded solution of(2.11)it is clear that

φ(y) = 1

g(∞)g(y),

sinceφ(∞) = 1.The object of this paper is to discuss the numerical solution of(2.7)–(2.9). Basically we will apply the Simp-

son rule which is known to have an error of order 4, provided the relevant functions areC4[0,∞), whereby Cq[0,∞) we shall meanq times continuously differentiable functions. The following result complementsTheorem 2.1.

Theorem 2.4. Assume thatp > 0, r > 12σ

2R and thatF ∈ C4[0,∞). Assume also thatF (x)xη stays bounded for

someη > 0.Then the survival probabilityφ satisfies(2.6)andφ ∈ C4[0,∞).

Proof. If σ2P + σ2

R = 0 orσ2P > 0 it follows from(2.8) and (2.9)together withTheorem 3.1in Linz (1985)that the

solution of(2.11)is in C4[0,∞). Whenσ2P = 0 andσ2

R > 0 it is shown inGaier and Grandits (2004)(henceforthG&G) thatg is inC2[0,∞), but the proof in their Theorem 5 is easily extended toC4[0,∞) as follows. Note firstthat upon replacingφ by g, then as in G&G, Eq.(2.6)can be written as

12(σ2

Ry2 + σ2

P )g′′(y) + (ry + p)g′(y) + λf (u)(y) = 0, (2.12)

where

f (u)(x) = 2λ

σ2R

(g(0)F (x) +

∫ x

0g′(y − z)F (z) dz

).

G&G solve(2.12)for g1 = g′ to get

g1(y) =(∫ y

0f (u)(x)xγ−2e−α/x dx

)y−γeα/y, (2.13)

whereα = 2pσ2R

andγ = 2rσ2R

> 1. Forq a positive integer,

(∫ y

0xβxγ−2e−α/x dx

)y−γeα/y = yβ

∫ ∞

0(1 + t)−(1+β+γ)e−(α/x)t dt = yβ

q∑i=1

ciyi + o(yβ+q),

where the first equality is just the substitutiont = xy

− 1 and the second is Watsons lemma(Davies, 1978). We can

now as in G&G argue recursively thatg1 ∈ Ci[0,∞) for i = 2,3, henceg ∈ C4[0,∞).If we can prove that there is an increasing bounded solution of(2.12)with g(0)> 0 if σ2

P = 0 andg(0) = 0 ifσ2P > 0, then it is clear thatφ(y) = g(y)

g(∞) will satisfy the boundary conditions(2.5), and hence byTheorem 2.1, φ

is the survival probability function. For the case withσ2P = 0 andσ2

R > 0 it is shown in G&G thatg is increasingand bounded providedF (x)x1+η is bounded for someη > 0. Their method would also work for the other cases as


well, but here we will present a completely different proof that may be of interest in itself. It also works under theweaker condition thatF (x)xη is bounded, thus making the theorem valid for, e.g. Pareto distributions without finiteexpectations.

It follows from (2.13)thatg′(y) > 0 for allywhenσ2P = 0 andσ2

R > 0. Solving(2.12)w.r.t.g′ as in(2.13)showsthatg′(y) > 0 for all y for the other cases as well. Therefore, to prove thatg is bounded, it is sufficient to prove that∫ y

0g(x) dx < cy(1 + ε(y)) (2.14)

for some constantc and a functionε satisfying limy→∞ ε(y) = 0. By the version of a Tauberian theorem given inLinz (1985), Theorem 6.8,(2.14)follows provided we can prove that

limt↓0

tg(t) = C (2.15)

for some constantC. Here

g(t) =∫ ∞

0e−tyg(y) dy

is the Laplace transform ofg. Now for anyy0 > 0 there is ak so thatK(y, x) ≤ k for all x ≥ 0 and ally ≥ y0. ByExample 3.3 inLinz (1985), g(y) ≤ c1eky for some positivec1, and therefore ˜g exists fort > k. Taking the Laplacetransform on both sides of(2.12)then gives fort > k,

12σ

2Rt

2g′′(t) + (2σ2R − r)tg′(t) + α(t)g(t) = β, (2.16)

where

α(t) = σ2R − r + pt + 1

2σ2P t

2 − λt ˜F (t), β = pg(0) + 12σ

2Pg

′(0), ˜F (t) = ∫ ∞0 e−tyF (y) dy.

Replacingt by a complexz, it is standard that the solution of(2.16)is analytical onRe(z) > 0, hence by Theorem10.1 in Chapter 5 ofWidder (1971), g(t) exists for allt > 0. Furthermore,t ˜F (t) = o(tη/2), so for the asymptoticsof g near 0 we may replaceα(t) by its dominating termσ2

R − r since this gives

limt↓0

tg(t) = limt↓0

tg(t),

whereg solves

12σ

2Rt

2g′′(t) + (2σ2R − r)tg′(t) + (σ2

R − r)g(t) = β. (2.17)

The general solution of(2.17)whenσ2R = r is

g(t) = c1t−1 + c2t

−ρ + β

σ2R − r

, (2.18)

wherec1 = 0 andc2 = 0 if and only ifσ2R > 0 in which caseρ = 2 − 2r

σ2R

< 1. Whenσ2R = r, the solution is

g(t) = c1t−1 + c2 + 2β

rln t. (2.19)

Eqs.(2.18) and (2.19)together now gives(2.15), and the theorem is proved.


Remark. Due to the smoothing effects in the model(2.2), particularly if σ2P > 0, theCq[0,∞) properties of

F are probably not necessary forTheorem 2.1to hold, see e.g. Proposition 3.4 andTheorem 3.2in Paulsen(1993)for a justification of this conjecture. However, they are necessary for the Simpson rule to have an error oforder 4.

3. Numerical methods

In this section we will discuss numerical solutions of the survival probabilityφ(y) using a fixed gridy =0, h,2h, . . . . It is assumed throughout that the assumptions ofTheorem 2.1hold. For this to be the case, byTheorem 2.2, it is necessary thatr > 1

2σ2R. The numerical solution will be of the form

gn + h

n∑i=1

wiKn,igi = βn, (3.1)

wheregi is the numerical approximation tog(ih), Kn,i = K(nh, ih) andβn = β(nh). Thewi are the integrationweights. Here we shall use the block-by-block method in conjunction with the Simpson integration rule, thusobtaining solutions in blocks of 2. To briefly explain the method, Simpsons rule gives for anyk ∈ C4[ih, (i+ 2)h],

∫ (i+2)h

ih

k(x) dx = h

3(k(ih) + 4k((i+ 1)h) + k((i+ 2)h)) +O(h5).

Therefore, forn = 2, (3.1)becomes

g2 + 13h(K2,0g0 + 4K2,1g1 +K2,2g2) = β2. (3.2)

Hereg1 is unknown, but using the same rule with stepsizeh/2, Simpsons rule gives

g1 + 16h(K1,0g0 + 4K1,(1/2)g(1/2) +K1,1g1) = β1. (3.3)

Quadratic interpolation gives thatg(1/2) ≈ 38g0 + 3

4g1 − 18g2 and inserting this into(3.3)yields

g1 + 16h

((K1,0 + 3

2K1,(1/2)

)g0 + (K1,1 + 3K1,(1/2))g1 − 1

2K1,(1/2)g2

)= β1. (3.4)

Eqs.(3.2) and (3.4)is a pair of equations to solve forg1 andg2. Continuing like this in blocks of 2 we get,

g2m+2 + h

2m∑i=0

wiK2m+2,igi + h

3(K2m+2,2mg2m + 4K2m+2,2m+1g2m+1 +K2m+2,2m+2g2m+2) = β2m+2

(3.5)

and

g2m+1 +h2m∑i=0

wiK2m+1,igi+ h

6(K2m+1,2mg2m + 4K2m+1,2m+(1/2)g2m+(1/2) +K2m+1,2m+1g2m+1) = β2m+1

(3.6)

Approximatingg2m+(1/2) by 38g2m + 3

4g2m+1 − 18g2m+2 and inserting this into(3.5) yields a set of two linear

equations forg2m+1 andg2m+2.In order to solve for thegn, a starting value is required. Whenσ2

P = 0, any positive value ofg0 can be chosen.Whenσ2

P > 0, it follows from Theorem 2.1that g0 = 0 must be used, whileg′(0)> 0 can be chosen arbitrary.


This is because in both cases different starting values will cancel inφ(y) = g(y)/g(∞). This was confirmed by thenumerical examples, where we found that the results did not depend on the level of these starting values.

Theorem 3.1. Given the assumptions ofTheorem 2.3,assume that(2.11)is solved using the block-by-blockmethodin conjunction with Simpsons rule. Let y be fixed so thaty = nh. If σ2

P > 0 then withg0 = 0 andg′(0)> 0,

|gn − g(y)| = O(h4). (3.7)

If σ2P = σ2

R = 0, then(3.7)also holds for anyg0 = g(0)> 0. If σ2P = 0andσ2

R > 0, thenwith anyg0 = g(0)> 0,(3.7)holds forn = 0 andn > 2 provided that forn = 1,2,

gn =(

1 + λ

pnh+ λ(λ− r) − 2f (0)λp

p2 (nh)2)g0

and the block-by-block method is used forn > 2.Furthermore, if y is chosen so that

|g(y) − g(∞)| = O(hq0), (3.8)

then withn0 = y/h and

φn = gn

gn0

,

the difference(for n = 0 andn > 2 if σ2P = 0 andσ2

R > 0)

|φn − φ(y)| = O(hq),

with q = min4, q0.

Proof. Except for the case withσ2P = 0 andσ2

R > 0, it follows from results in Chapter 7 ofLinz (1985)that thesolution satisfies(3.7)providedg is four times continuously differentiable as is the case here byTheorem 2.3.

The case whenσ2P = 0 andσ2

R > 0 needs special consideration at 0 due to the (removable) singularity there.Writing g(y) = γ0 + γ1y + γ2y

2 +O(y3) we shall see below how to identifyγ1 andγ2. Clearlyγ0 = g(0). Thensettinggi = γ0 + γ1ih+ γ2(ih)2 gives that|gi − g(ih)| = O(h3) for i = 1,2. Usingg0, g1 andg2 as starting valuesin the block-by-block method, it follows from e.g. formula (7.20) inLinz (1985)that(3.7)holds forn > 2 in thiscase as well.

To identifyγ1 andγ2, write

K(y, x) = 2

σ2Ry

2(κ0 + κ1x+ κ2F2(y − x) − κ3y),

where

κ0 = p, κ1 = 2r − 3σ2R + λ, κ2 = λ, κ3 = r − σ2

R + λ.


ApproximatingF2(x) by f2x2 + f3x

3 (sinceF2(0) = F ′2(0) = 0), calling the resultingK(y, x) by K(y, x), we

get

g(y) +∫ y

0K(y, x)g(x) dx = β(y) +O(y2) (3.9)

whereg(y) = γ0 + γ1y + γ2y2. Identifying constant terms as well as terms of first order iny then gives

γ0 + 2

σ2R

(1

2κ0γ1 + 1

2κ1γ0 − κ3γ0

)= 0 (3.10)

and

γ1 + 2

σ2R

(1

3κ0γ2 + 1

3κ1γ1 + 1

3f2κ2γ0 − 1

2κ3γ0

)= 0. (3.11)

Solving (3.10)for γ1 in terms ofγ0 and then inserting this value into(3.11)to find γ2 give after inserting theproper values of theκi’s,

g(y) =(

1 + λ

py + λ(λ− r) − 2f (0)λp

p2 y2)g(0) +O(y3), (3.12)

where we used thatf2 = 12F

′′2 (0) = 1

2f (0) with f the density ofF.Finally,

|φn − φ(y)| =∣∣∣∣ gngn0

− g(y)

g(∞)

∣∣∣∣ =∣∣∣∣ gn

g(∞)

1

1 + ((gn0 − g(∞))/g(∞))− g(y)

g(∞)

∣∣∣∣= 1

g(∞)|gn − g(y)| +O(hq) = O(hq)

since

|gn0 − g(∞)| ≤ |gn0 − g(y)| + |g(y) − g(∞)| = O(h4) +O(hq0) = O(hq).

This ends the proof of the theorem.

The condition(3.8) is a bit impractical since it only concerns the order of convergence and not the accuracy ofthe convergence itself. What we really want is that|g(y) − g(∞)| is negligible compared to|gn − g(y)|. This willbe the case if|g(y) − g(∞)| is of the same size as the numerical accuracy of the computedgn (due to standardtruncation errors depending on the number of digits used by FORTRAN, see the discussion inExample 4.1). Sowhat we typically do in practice is to try a ¯y and see if thegn stabilize asn approachesn0 = y/h. If they do,y islarge enough, but otherwise ¯y has to be increased until stability is obtained. To this end it may be useful to knowthat for the block-by-block method|g2m+2 − g2m+1| = O(h4), hence these oscillations will make it slightly moredifficult to judge when stability is obtained.

Nevertheless, some asymptotics can be indicative of how large ¯y should be. Note that

|g(y) − g(∞)| = g(∞)ψ(y),

and using FORTRAN with DOUBLE PRECISION, experience has shown us that the computational accuracy isabout 10−γ whereγ will vary between 8 and 10 withγ closer to 8 ifn0 is large, see Section4. Therefore, ify is


chosen so that

ψ(y) ≈ 10−γ , (3.13)

theny should be sufficiently large.In order to find if ay can be found so that(3.13)is satisfied, the asymptotic properties ofψ(y) can sometimes be

useful. A lot is known about these properties, and here is a rough, but for our purpose sufficient accurate descriptionof the asymptotics. Again it is as always assumed thatr > 1

2σ2R. By a slowly varying functionl we mean that

limx→∞ l(tx)/l(x) = 1 for all t > 0.Assume first thatσ2

R = 0 in which case the asymptotics depend strongly onF. If for somec > 0, F (y)ecy goes tozero, thenψ(y) goes to zero exponentially fast. In this case ¯y do not have to be very large, but some experimentationalong the lines described above is recommended.

Consider now the following 3 heavy tailed cases, which will be sufficient for our purpose.

Case 1. σ2R = 0 andF (y) ∼ y−κl(y) for some slowly varyingl. Then

ψ(y) ∼ λ

κry−κl(y).

Case 2. σ2R > 0 and limy→∞ yκ1F (y) = 0 whereκ1 = 2r

σ2R

− 1. Then

ψ(y) ∼ cy−κ,

whereκ = κ1 andc is a constant.

Case 3. σ2R > 0 andF (y) ∼ y−κl(y) for some slowly varyingl whereκ < κ1 with κ1 given as inCase 2above.

Then

ψ(y) ∼ 2λ

κσ2R(κ1 − κ)

y−κl(y).

For more details on these and other cases withσ2R = 0, seeAsmussen (1998), andKl uppelberg and Stadtmuller

(1998)and the survey paper byPaulsen (1998a). For more information onCases 2 and 3, see e.g.Frolova et al.(2002), Kalashnikov and Norberg (2002), Paulsen (2002), andGaier and Grandits (2004).

For all these three heavy tailed cases,ψ(y) ∼ cy−κl(y) whereκ andl(y) are known, whilec is known inCases 1and 3. There is a formula forc in Case 2as well, but this is useless in practice. Then(3.13)is satisfied provided

y = (10γcl(y))1/κ. (3.14)

Formula(3.14)is just meant to give a rough idea of the size of ¯y. Checking if thegn stabilize asn approachesn0 = y/h is still recommended. When ¯y must be chosen very high in order for|g(y) − g(∞)| to be sufficientlysmall, a method to avoid the calculation of all thegn for n = 1, . . . , n0 is proposed in Section4.3. This methodmakes use of the above asymptotics.

As mentioned above, for the block-by-block method,|g2m+2 − g2m+1| = O(h4) as well, so in order to limit theeffect of these oscillations, let

g∞ = 1

m0 + 1

n0∑k=n0−m0

gk, (3.15)


where againn0 = y/h. As just discussed, ¯y must be large enough so that|g(y −m0h) − g(∞)| is negligible. Thenwe set

φn = gn

g∞. (3.16)

It may well be that analytical expression forF (x) and

F2(x) = xF (x) −∫ x

0vf (v) dv

do not exist, in which case they must be calculated numerically. Typically this gives values only on grid pointsx = 0, 1

2h, h,32h, . . . yielding numeric valuesF (ih/2) andF2(ih/2). Note from(3.6) that values are required on

multiples ofh/2. However, interpolating between grid points yields functionsF andF2 that are inCq[0,∞) foranyq ≥ 0 and so that

supy

(|F (y) − F (y)| + |F2(y) − F2(y)|) < Chm (3.17)

for someC > 0. Herem is the order of the error when numerically calculatingF andF2. If the Simpson rule isused thenm = 4. Doing so gives instead of(2.11),

g(y) +∫ y

0K(y, x)g(x) dx = β(x),

whereK is the same asK, but with the appropriate ofF or F2 replaced byF or F2. Actually, since Simpsons rulerequires an odd number of gridpoints, to calculate, e.g.F (ih/2) we need a gridsize ofh/4 since then the numericalrecursion of e.g.

F

(ih

2

)= F

((i− 1)

h

2

)+

∫ ih/2

(i−1)h/2f (x) dx

becomes

F

(ih

2

)= F

((i− 1)

h

2

)+ h

12

(f

((i− 1)

h

2

)+ 4f

((i− 1

2

)h

2

)+ f

(ih

2

)). (3.18)

Theorem 3.2. In addition to the assumptions ofTheorem 3.1, assume that F orF2 are numerically calculatedusing Simpsons rule as in(3.18). Let gn be the numerical value ofg(nh) using the same method as described abovefor calculatinggn. Then the results ofTheorem 3.1still hold withgn replaced bygn andφn replaced by

φn = gn

gn0

.

Proof. We have

|gn − g(nh)| ≤ |gn − g(nh)| + |g(nh) − g(nh)|.


Interpolating theF (nh) or theF2(nh) so that they satisfy the assumptions ofTheorem 2.1shows just as in theabove discussion of|gn − g(nh)| that|gn − g(nh)| = O(h4). It therefore remains to prove that

|g(nh) − g(nh)| = O(h4). (3.19)

If σ2P + σ2

R = 0 or σ2P > 0, using an interpolation that satisfies(3.17), theresult (3.19)follows from Theorem

3.10 inLinz (1985). It therefore only remains to prove(3.19) for the caseσ2P = 0 andσ2

R > 0. Again we use an

interpolation that satisfies(3.17)and so thatF2 ∈ C4[0,∞). In addition it can be made to satisfy supy | d4

dy4 F2(y)| <C1 for someC1 > 0. Since|F2(x) − F2(x)| = O(h5) for x ≤ h andF2(0) = F ′

2(0) = 0, it follows that we in thiscase can assume

F2(x) =4∑i=2

fixi +O(h5), F2(x) =

4∑i=2

fixi +O(h5). (3.20)

Therefore, writingg(y) = ∑4i=0 γix

i +O(h5) andg(y) = ∑4i=0 γix

i +O(h5), it follows using the same argu-ments as in(3.9)–(3.12)together with(3.20)thatγi = γi for i ≤ 4. This is because, e.g.γ4 only depends onfi, i ≤ 4,but by(3.20)for thesei, fi = fi. Therefore,

|g(y) − g(y)| = O(h5), 0 ≤ y ≤ h.

If we can prove that for anyi andy ≤ h,

|g(ih+ y) − g(ih) − (g(ih+ y) − g(ih))| = O(h5), (3.21)

then summing up gives(3.19).We will prove(3.21)by induction, so assume it holds fori = 0,1, . . . , k. Then for 0≤ y ≤ h, some calculations

give

vk(y) +∫ y

0K(kh+ y, kh+ x)vk(x) dx = βk(y),

where

vk(y) = g(kh+ y) − g(kh)

and

βk(y) = β(kh+ y) − β(kh) −k∑i=1

g((i− 1)h)∫ ih

(i−1)h(K(kh+ y, x) −K(kh, x)) dx

−k∑i=1

∫ ih

(i−1)h(K(kh+ y, x) −K(kh, x))(g(x) − g((i− 1)h)) dx− g(kh)

∫ kh+y

kh

K(kh+ y, x) dx.

As an example, we show that∣∣∣∣∣g(kh)∫ kh+y

kh

K(kh+ y, x) dx− g(kh)∫ kh+y

kh

K(kh+ y, x) dx

∣∣∣∣∣ = O(h4). (3.22)


To show this, note first that by the induction hypothesis|g(kh) − g(kh)| = O(h4). Furthermore withc = 2λσ2R

,

∣∣∣∣∣∫ kh+y

kh

(K(kh+ y, x) − K(kh+ y, x)) dx

∣∣∣∣∣ ≤ c

(kh+ y)2

∣∣∣∣∣∫ kh+y

kh

(F (kh+ y − x) − F (kh+ y − x)) dx

∣∣∣∣∣≤ c

k2h2

∣∣∣∣∫ y

0(F (y − x) − F (y − x)) dx

∣∣∣∣ = 1

k2h2O(h6) = 1

k2O(h4).

This proves(3.22). Similar arguments using that the induction hypothesis also allows us to write|g((i− 1)h) −g((i− 1)h)| = (i− 1)O(h5), show that in fact for 0≤ y ≤ h,

βk(y) = βk(y) +O(h4),

whereβk is the same asβk, but withgandK replaced by ˆg andK, respectively. We therefore get, again for 0≤ y ≤ h,

vk(y) +∫ y

0K(kh+ y, kh+ x)vk(x) dx = βk(y) +O(y4) (3.23)

and by definition

vk(y) +∫ y

0K(kh+ y, kh+ x)vk(x) dx = βk(y). (3.24)

Comparing(3.23) and (3.24)with (3.9) and (2.11)and using the same arguments as after(3.9), but with two moreterms in the expansion ofvk(y) andvk(y), it follows from(3.20)and the discussion following it that|vk(y) − vk(y)| =O(h5). This ends the induction hypothesis and hence the proposition is proved.

Remark. In practice we did as in(3.15)and calculated

g∞ = 1

m0 + 1

n0∑k=n0−m0

gk,

setting

φn = gn

g∞(3.25)

in accordance with(3.16).

4. Numerical results

In this section we will report some numerical results obtained using the methods described in Section3. For agiveny, letψ(y) = 1 − φ(y) be the ruin probability, and letψh(y) be the calculated ruin probability using(3.16)when a gridsizeh is used, and the relevant ofF orF2 is calculated analytically. Similarly, using(3.25), letψh,h0(y)be the same asψh(y), but with the relevant ofF or F2 calculated using the Simpson rule with gridsizeh0/4 asdescribed in(3.18). Hereh/h0 must be an integer. In all the calculations ofg∞ or g∞,m0 = 1000 which is probablyunnecessary high.


Whenψ(y) is known, we set

Dh(y) = 100× ψh(y) − ψ(y)

ψ(y)and Dh,h0(y) = 100× ψh,h0(y) − ψ(y)

ψ(y),

i.e. the percentage relative error.In the examples two different distributions forS(generic for theSi) are used. The first is the standard exponential

distribution, i.e.

f (x) = e−x, x > 0, (4.1)

so that in particularE[S] = 1. The other is the Pareto distribution

f (x) = (α− 1)αα(α− 1 + x)−(1+α), x > 0, α > 1, (4.2)

and againE[S] = 1. Furthermore with this distribution,

F (x) =(

α− 1

α− 1 + x

)α

and

F2(x) = x− 1 +(

α− 1

α− 1 + x

)α−1

.

Note that asymptotically

F (x) ∼ (α− 1)αx−α = l(x)x−κ,

where thenl(x) is the constant (α− 1)α andκ = α.All the calculations were done on a PC with an AMD Athlon XP 3200+ processor and 1024 MB internal memory.

The program language was FORTRAN, and we used DOUBLE PRECISION to get satisfactory accuracy. Of courseslower programs like Splus, R, Gauss, Matlab, Maple or Mathematica could have been used, but only at the expenseof a considerably longer computing time.

The calculations go to an upper limit ¯y, the choice of which is discussed in the examples. Hencen0 = y/h gi’sare required, and the number of elementary calculations for these go liken2

0. It turned out that the computing timefor n0 = 10,000 is 4 s (yes, it is really fast), hence forn0 = 100,000 it is about 7 min and forn0 equal to 1 millionit is about 11 h.

On the other hand, the calculation of theF (ih) or theF2(ih) is only linear in the number required, hence letting,e.g.h0 = h/10 when using the Simpson rule really does not add much to the computing time.

4.1. The case without diffusion

Dickson and Waters (1999)presented several numerical methods for this case, but without any effort to analyzethe error rate. Of all the methods they discuss, the one that is closest to that used in this paper is a method proposedby Sundt and Teugels (1995). They showed numerically that it performs very well, although it is clear from thealgorithm of this method that its error is of order no higher than 2. On the other hand, the Sundt and Teugels methodhas the advantage that it provides an upper and lower bound, which of course is useful unless these bounds are toofar from each other.

The following example shows that the block-by-block method performs extremely well.


Table 1Ruin probabilities and relative errors

y ψ(y) ψ0.1(y) D0.1(y) ψ0.01(y) D0.01(y)

0 0.7909540044 0.7909543928 0.0000 0.7909540044 0.00005 0.1776111024 0.1776112352 0.0001 0.1776111048 0.0000

10 0.0241449177 0.0241449299 0.0001 0.0241449168 0.000015 0.0022199914 0.0022199735 0.0008 0.0022199906 0.000020 0.0001502219 0.0001502021 −0.0132 0.0001502171 −0.003225 0.0000079593 0.0000079395 −0.2484 0.0000079565 −0.035630 0.0000003456 0.0000003237 −6.3518 0.0000003399 −1.656435 0.0000000127 0.0000000013 −89.4923 0.0000000130 1.822340 0.0000000004 −0.0000000004 −203.0081 0.0000000103 2419.8958

Table 2Relative error for various values ofh

y D1(y) D0.5(y) D0.1(y) D0.05(y) D0.01(y) D0.005(y)

0 0.0026 0.0002 0.0000 0.0000 0.0000 0.00005 1.1428 0.0140 0.0001 0.0000 0.0000 0.0000

10 −4.5876 −0.1513 0.0001 0.0000 0.0000 0.000015 25.556 −1.9577 −0.0008 0.0000 0.0000 −0.000120 −657.32 −26.432 −0.0132 −0.0035 −0.0032 −0.004325 5346 −446.41 −0.2484 −0.0311 −0.0356 −0.009730 −227627 −9273 −6.3518 −1.8852 −1.6564 −2.066535 2759602 −229115 −89.492 −25.475 1.8223 9.4278

Example 4.1. In Tables 1–4, the valuesp = 1.1, λ = 1, r = 0.05 are used, and in additionS is assumed to beexponentially distributed with expectation 1 as in(4.1). For this model the true ruin probability was found way backby Segerdahl (1942)and later reappeared in many papers. This enables us to assess the quality of the numericalmethod, just as was done byDickson and Waters (1999). The exact ruin probability is calculated to a high degreeof accuracy from Segerdahl’s formula using the Splus programintegrate.

It is seen fromTable 1that the results are extremely good. Lettingh = 0.1 gives accurate results down to truevalues of about 10−6 and whenh = 0.01 they go all the way down to about 10−8. For lower ruin probabilities thanthat, i.e. wheny = 40, so thatψ(y) = 0.4 × 10−10, the number of digits used by FORTRAN is not sufficient, andthe result becomes poor. For some reason the relative error withh = 0.01 is about the same wheny = 30 as wheny = 35, which is a bit strange. Trying various values of ¯y did not change this unreasonable result. In the tables,y = 100 was used.

Table 2shows the relative percentage error for various stepsizesh. We see that there is a steady gain ashdecreases fromh = 1 down toh = 0.05. After that the picture is somewhat less clear, and in particular when goingfrom h = 0.01 to 0.005 the error increases for mosty values. The reason for this is probably that due to the limited

Table 3Comparing exact values against those obtained using Simpsons rule for the claim distributions

y D0.1(y) D0.1,0.1(y) D0.1,0.01(y) D0.01(y) D0.01,0.01(y)

0 0.0000 0.0000 0.0000 0.0000 0.00005 0.0001 0.0000 0.0001 0.0000 0.0000

10 0.0001 −0.0007 0.0001 0.0000 0.000015 −0.0008 −0.0083 −0.0008 0.0000 0.000020 −0.0132 −0.1177 −0.0132 −0.0032 −0.003225 −0.2484 −2.0262 −0.2485 −0.0356 −0.035730 −6.3518 −44.870 −6.3528 −1.6564 −1.657135 −89.492 −1177 −89.510 1.8223 1.8049


Table 4Comparing single and double precision

y D0.1(y) DSP0.1(y) DSP0.1,0.1(y) DSP0.1,0.01(y) D0.01(y) DSP0.01(y) DSP0.01,0.01(y)

0 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.00005 0.0001 0.0006 0.0213 0.1188 0.0000 0.0002 0.0601

10 0.0001 0.0046 0.1834 1.0531 0.0000 0.0003 0.542715 −0.0008 0.0537 1.9844 11.820 0.0000 0.0090 6.160420 −0.0132 0.8604 28.126 168.55 −0.0032 −0.1346 84.71025 −0.2484 16.471 504.87 3011 −0.0356 44.194 143330 −6.3518 390.21 11031 65837 −1.6564 682.83 2951335 −89.492 11642 285622 1702136 1.8223 −195296 707495

number of digits used by FORTRAN, the gain of decreasing the stepsize is offset with the loss of having to calculatetwice as manygi’s for the samey-value.

In Table 3, the effect of using the Simpson rule to calculateF (y) instead of using the exact value is considered. Wesee that withh0 = 0.01 there is virtually no effect, which is really good news. Whenh0 is only 0.1 (andh = 0.1),there is some loss in quality compared to when using exact values, but as mentioned in the introduction to thissection, decreasingh0 from h to h/10 does not increase the computing time by much.

Initially we used the single precision REAL in the FORTRAN programs, but the results were not quite satisfactory,as we can see fromTable 4. Here the superscriptSP indicates that single precision is used. It is seen that with singleprecision, accurate results are only obtained forψ(y) as large as 10−4, and furthermore that using Simpson’s rule tocalculateF (y) resulted in considerably poorer results, even whenh0 = 0.01. Another consequence of using singleprecision is that the calculated ruin probabilities depend much more on the choice of ¯y due to fairly large oscillationsin thegn.

4.2. The case with diffusion

In all the examples in this subsection, the valuesp = 1.1, λ = 1, r = 0.1 andσR = 0.2 are fixed. This givesκ1 = 2r

σ2R

− 1 = 4.

Example 4.2. In this exampleS is exponentially distributed with expectation 1 andσP = 0.2. We see fromTable 5that except for the caseψ0.1,0.1(y), the results are for any practical purpose identical. This isCase 2discussed afterTheorem 2.1. Sincec is unknown, formula(3.14) is of limited use, but lettingc = 1 andγ = 10givesy = 102.5 ≈ 316. InTable 5, y = 1000 was chosen. A little experimentation showed thatg∞ changed almostnothing whenever ¯y was at least 600, giving confidence in the results. This is in contrast to using single precision,where as mentioned at the end ofExample 4.1, the calculatedg∞ would oscillate quite markedly with ¯y.

Table 5Calculated ruin probabilities with exponentially distributed jumps andσP = 0.2

y ψ0.1(y) ψ0.1,0.1(y) ψ0.1,0.01(y) ψ0.01(y) ψ0.01,0.01(y)

5 0.16875282 0.16875271 0.16875282 0.16875159 0.1687515910 0.03804299 0.03804287 0.03804299 0.03804278 0.0380427820 0.00390649 0.00390639 0.00390649 0.00390644 0.0039064450 0.00010944 0.00010936 0.00010944 0.00010946 0.00010946

100 0.00000676 0.00000670 0.00000676 0.00000676 0.00000676200 0.00000042 0.00000038 0.00000042 0.00000041 0.00000041300 0.00000008 0.00000005 0.00000008 0.00000008 0.00000008


Table 6Calculated ruin probabilities with Pareto distributed jumps withα = 2 andσP = 0.2

y ψ0.5(y) ψ0.5,0.05(y) ψ0.5,0.01(y) ψ0.2(y) ψ0.1(y)

10 0.09922014 0.09921874 0.09922014 0.09887592 0.0988668950 0.00559161 0.00559033 0.00559161 0.00557826 0.00557795

200 0.00032490 0.00032388 0.00032489 0.00032409 0.00032407500 0.00005079 0.00004995 0.00005078 0.00005066 0.00005065

1000 0.00001260 0.00001191 0.00001259 0.00001256 0.000012562000 0.00000313 0.00000258 0.00000313 0.00000312 0.000003125000 0.00000049 0.00000013 0.00000049 0.00000049 0.00000049

Example 4.3. In this exampleShas the distribution(4.2)with α = 2, whileσP = 0.2 as inExample 4.2. This isCase 3in Section3, and(3.14)givesy = (10γ × 12.5)1/2. With γ = 10 this gives ¯y = 353,000. This is very high,but the more reasonable (since so manygn have to be calculated, increasing errors due to a limited number of digitsused by FORTRAN)γ = 8 givesy = 35,300. Indeed some experimentation withh = 0.5 showed that ¯y = 30,000is sufficient, and consequently that is the value we have used inTable 6. In fact, for all heavy tailed ruin probabilitiesit turned out that in order to experimentally find a sufficiently large ¯y, a fairly large value ofh was just as goodas a small one, hence a good ¯y could be obtained rather fast except from the exceptionally heavy tailed cases, seeExample 4.8below.

Again the results were stable w.r.t. changes in ¯y as long as ¯y was at least 30,000. It is seen from the table thatψ0.2(y) andψ0.1(y) are more or less exactly the same, and evenψ0.5(y) do not differ much. The fact that heavytailed ruin probabilities allowed for biggerh was another positive lesson learned from the numerical work. In thetable the values forψ0.2,0.01(y) andψ0.1,0.01(y) have been omitted, the reason for that is that they were exactly equaltoψ0.2(y) andψ0.1(y), respectively.

Example 4.4. HereS has the distribution(4.2) with α = 3 andσP = 0, so therefore the quadratic approxi-mation given inTheorem 3.1to getg1 andg2 was used. Here(3.14) gives y = (10γ 50

3 8)1/3. With γ = 10 thisgives y = 11006 whileγ = 8 gives y = 2371. The results with ¯y = 5000 are reported inTable 7, and againthey are stable and reasonable when compared to results obtained for higher ¯y values. In the last column,the ψ0.1(y) are the calculated ruin probabilities with the same parameters except thatσP has been changed toσP = 0.01 and the standard block-by-block method without using the quadratic approximation to getg1 andg2 was used. This column then serves as a control of the quadratic approximation, and we see that exceptfrom the obvious case wheny = 0, the two methods give virtually the same result. Ruin probabilities withσP = 0.01 are always the highest, as should be. InTable 8, ruin probabilities for the two casesσP = 0 and0.01 are compared for smally values, and it is seen that the effect of changingσP from 0 to 0.01 dies outrapidly.

Table 7Calculated ruin probabilities with Pareto distributed jumps withα = 3 andσP = 0

y ψ0.5(y) ψ0.1(y) ψ0.1,0.01(y) ψ0.02(y) ψ0.1(y)

0 0.73416167 0.73421098 0.73421098 0.73421112 1.0000000020 0.01657526 0.01652787 0.01652787 0.01652783 0.0165283250 0.00124534 0.00123814 0.00123814 0.00123814 0.00123818

100 0.00015034 0.00014858 0.00014858 0.00014858 0.00014859200 0.00001816 0.00001772 0.00001772 0.00001772 0.00001772500 0.00000116 0.00000109 0.00000109 0.00000109 0.00000109

1000 0.00000015 0.00000013 0.00000013 0.00000013 0.00000013


Table 8Comparison betweenσP = 0 and 0.01 for smally-values

y = 0 y = 1 y = 2 y = 3 y = 5

ψ0.1(y) 0.73421098 0.54716895 0.42102326 0.32878118 0.20686509ψ0.1(y) 1.00000000 0.58155668 0.42106264 0.32880088 0.20687433

4.3. Using asymptotics for very small and very heavy tailed ruin probabilities

In the examples so far we have only made limited use of the asymptotic results stated in Section3. We saw inExample 4.1that the number of digits used by FORTRAN is not sufficient to get ruin probabilities much below10−8 (which probably is enough anyway). Also, if the ruin probability is very heavy tailed, it may be difficult tocomputeg∞.

In this section we shall assume thatψ(y) ∼ cy−κl(y) wherel is a slowly varying function andc is a constant. Thisasymptotics is discussed in Section3, wherec andl are explained. In particularl is known sinceF (y) ∼ l(y)y−κ.It is numerically the most challenging since otherwise the ruin probabilities will go sufficiently fast to zero for themethods already discussed to work well.

We start with the problem of computing very small ruin probabilities. Fory1 andy sufficiently large,

ψ(y) ≈(y

y1

)−κl(y)

l(y1)ψ(y1). (4.3)

We therefore define

ψh,y1(y) =(y

y1

)−κl(y)

l(y1)ψh(y1), y ≥ y1, (4.4)

where in particularψh,y1(y1) = ψh(y1).

Example 4.5. This is the same asExample 4.2, and the results are reported inTable 9. The columnψ0.1(y) is takenfrom Table 5, and theψ0.1,y1(y) are calculated according to(4.4), using that hereκ = 4 andl(y) is constant.

Comparingψ0.1(y) with the ψ0.1,y1(y) for different y and y1, we see that the asymptotics(4.3) works verywell, particularly wheny1 ≥ 50. Going below the horizontal line to highery-values, using e.g.ψ0.1,300(2000)=4.20e−11= 4.20× 10−11 seems to be a good approximation toψ(2000).

Table 9Comparison of exact and asymptotically calculated ruin probabilities

y ψ0.1(y) ψ0.1,20(y) ψ0.1,50(y) ψ0.1,100(y) ψ0.1,200(y) ψ0.1,300(y)

20 0.00390649 0.0039064950 0.00010944 0.00010000 0.00010944

100 0.00000676 0.00000625 0.00000680 0.00000676200 0.00000042 0.00000039 0.00000043 0.00000042 0.00000042300 0.00000008 0.00000008 0.00000008 0.00000008 0.00000008 0.00000008

500 1.000e−8 1.094e−8 1.081e−8 1.076e−8 1.037e−81000 6.25e−10 6.84e−10 6.76e−10 6.70e−10 6.72e−102000 3.90e−11 4.27e−11 4.22e−11 4.19e−11 4.20e−11


Table 10Estimatedκ-values fromExample 4.2

(y1, y2) (5,10) (10,20) (20,50) (50,100) (100,200) (200,300)

κ0.1(y1, y2) 2.149 3.284 3.902 4.017 4.011 3.993


(y1, y2) (20,50) (50,100) (100,200) (200,500) (500,1000)

κ0.1(y1, y2) 2.828 3.059 3.068 3.039 3.019

Another, but of course related way to assess whether the asymptotics(4.4) will work is to compare the trueκwith an estimated value. To be more specific, note that(4.3)gives

κ ≈ ln(ψ(y1)/ψ(y)) + ln(l(y)/l(y1))

ln(y/y1).

Therefore, fory1 < y2 we define

κh(y1, y2) = ln(ψh(y1)/ψh(y2)) + ln(l(y2)/l(y1))

ln(y2/y1). (4.5)

Whenevery1 is large enough so that ˆκh(y1, y2) is close to the trueκ, approximatingψ(y) with ψh,y2(y) will be quiteprecise fory > y2 and sufficiently smallh.

Example 4.6. Table 10contains the ˆκ0.1(y1, y2) calculated fromExample 4.2andTable 11comes fromExample4.4. In Table 10κ = 4 while in Table 11κ = 3. In both casesl(y) is a constant. We see that the estimated valuesbecome close to the trueκ-values whenevery1 andy2 are large enough.

Table 12contains the estimated ˆκh(y1, y2) from the fairly heavy tailed model inExample 4.3whereκ = 2. Wesee that fory1 moderately large the estimates are again good. The only surprise is that the error in ˆκh(2000,5000)is larger than in ˆκh(1000,2000) say.

Another interesting and useful observation is that the ˆκh(y1, y2) are basically unchanged whenh goes from 0.2to 1, as opposed to theψh(y) which vary much more.

Finally we shall look more into the case with very heavy tailed ruin probabilities. Although we may be interested inψ(y) for small or moderately large values ofy, the formulaψ(y) = 1 − g(y)/g(∞) seems to demand the calculationof g(x) for large values ofx in order to have a good approximation ofg(∞). However,(4.3) gives fory1 andy2sufficiently large

1 − (g(y2)/g(∞))

1 − (g(y1)/g(∞))= ψ(y2)

ψ(y1)≈

(y2

y1

)−κl(y2)

l(y1).


(y1, y2) (10,50) (50,200) (200,500) (500,1000) (1000,2000) (2000,5000)

κ1.0(y1, y2) 1.793 2.052 2.025 2.011 2.007 2.016κ0.2(y1, y2) 1.786 2.053 2.025 2.011 2.008 2.015


Table 13Comparison of standard calculated values, calculated values using a low upper limit and adjusted calculated values

y ψ0.1(y) ψ0.1,2200(y) ψ0.1,1000,2000(y)

10 0.09886689 0.09886446 0.0988669050 0.00557795 0.00557527 0.00557797

200 0.00032407 0.00032138 0.00032408500 0.00005065 0.00004796 0.00005067

1000 0.00001256 0.00000986 0.000012572000 0.00000312 0.00000042 0.00000313

This gives

g(∞) ≈ g(y2) − (y1/y2)κ(l(y2)/l(y1))g(y1)

1 − (y1/y2)κ(l(y2)/l(y1)).

So withy = nh, define

ψh,y1,y2(y) = 1 − gn

g∞,y1,y2

, (4.6)

where

g∞,y1,y2 = gn2 − (n1/n2)κ(l(n2h)/l(n1h))gn1

1 − (n1/n2)κ(l(n2h)/l(n1h))

with n1h = y1 andn2h = y2.We have used the notationψh(y) for the numerical approximation toψ(y) when the upper limit ¯y is large enough

so thatψ(y −m0h) is practically equal to zero. When this is not the case, write

ψh,y(y) = 1 − gn

g∞,y

,

wherey = nh, g∞,y is as in(3.15), but nowy is not large enough forψ(y −m0h) to equal zero. This implies inparticular thatψh,y(y) < ψh(y). Also if we let κh,y(y1, y2) be as in(4.5), but withψh(yi) replaced byψh,y(yi), thenit is easy to show that

κh,y(y1, y2) > κh(y1, y2) (4.7)

wheneverl(y) is a constant.

Example 4.7. This is Example 4.3again. In addition to theψ0.1(y) taken fromTable 6, Table 13also containstheψ0.1,2200(y) and theψ0.1,1000,2000(y). As could be expected, setting ¯y = 2200 results inψ0.1,2200(y) values toofar fromψ(y) asy approaches 2000. However, using the adjusted value(4.6)with y1 = 1000 andy2 = 2000 givesalmost perfect results, so it is not necessary to go all the way to ¯y = 30,000 or even higher as was done inExample4.3.

Example4.8. In this examplep = 1.1, λ = 1, S has the distribution(4.2)withα = 2,r = 0.03 andσP = σR = 0.2.This is thenCase 2in Section3 with κ1 = 0.5 so that asymptoticallyψ(y) ∼ cy−1/2 for somec > 0, indeed a heavytailed ruin probability. Using(3.14)with γ = 8 and the unknownc set equal to 1 gives ¯y = 1016. Therefore, it iscomputationally impossible to numerically calculateψh(y) for any reasonably smallh (using increasing stepsizewith y may have worked), so it is necessary to make use of the approximation(4.6). In order to findy1, y2 so that



(y1, y2) (1000,4000) (4000,10000) (10000,20000) (20000,50000)

κ10,1e7(y1, y2) 0.5251 0.5441 0.5683 0.6102

Table 15Estimated ruin probabilities fromExample 4.8

y ψ2,1.2e7(y) ψ2,5e6,1e7(y) ψ0.2,52000(y) ψ0.2,20000,50000(y)

500 0.078541 0.080043 0.072357 0.0795861000 0.054997 0.056538 0.048801 0.0562144000 0.026654 0.028241 0.020445 0.02807820000 0.011013 0.012626 0.004798 0.01255350000 0.006364 0.007985 0.000147 0.007939106 0.004022 0.005646 0.005614107 0.000155 0.001785 0.001775108 0.000564 0.000561109 0.000178 0.000177

g∞,y1,y2 ≈ g(∞), we calculated ˆκ10,1e7(y1, y2) (where 1e7 = 107), and the result is given inTable 14. It may beobjected thath = 10 is a very large stepsize, but inTable 12we saw that the estimated ˆκ do not depend muchon the stepsize, this was also confirmed here trying various stepsizes. FromTable 14we see that ˆκ10,1e7(y1, y2) isincreasing away fromκ = 0.5 asy1, y2 increase, but by(4.7) this is to be expected. Therefore, we conclude thaty1 = 20,000 is sufficiently large for the asymptotics to be accurate. InTable 15results are reported for the two casesh = 2 and 0.2. For largey-values the numbers have been extrapolated as in(4.4), so in particular

ψ2,5e6,1e7(y) =( y

107

)−0.5ψ2,5e6,1e7(107), y ≥ 107

and

ψ0.2,20000,50000(y) =( y

50000

)−0.5ψ0.2,20000,50000(50000), y ≥ 50000.

It is seen that theψ are almost the same in both cases, although bothhand (y1, y2) differ. The case withh = 2 givesthe highest estimated ruin probability, but from, e.g.Table 6we find that this is natural. So in conclusion we feelconfident that the approximationψ0.2,20000,50000(y) is sufficiently accurate.

Acknowledgments

This work was supported by the NUFU project 33/02. We also thank an unknown referee for several usefulsuggestions.

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A Numerical Method to Find the Probability of Ultimate Ruin

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